I have a fun problem for you : Let G be a simple graph with 3n² vertices (n≥2). It is known that the degree of each vertex of G is not greater than 4n, there exists at least a vertex of degree one, and between any two vertices, there is a path of length ≤3. Prove that the minimum number of edges that G might have is equal to( 7n² - 3n)/2. Hope you find it interesting.
Thanks for watching and I am not sure what you mean. Assuming that G and M are sets, A U (G\M) is just a set, so what is the problem? What is being asked? Do we know what A, G, and M are, and want to find A U (G\M)?
Thanks for watching! That would be the empty set - try to prove it! The intuition is that since A is a subset of B, it contains less than B, which means A' contains MORE than B', so taking A' away from B' would get rid of everything, leaving the empty set.
Let us use inductive reasoning and start backward from the end result. Let us assume that A = {1, 2, 3} and B = {1, 2, 4}. Therefore A - B = {s | s belongs to A and s does not belong to B} = {3}. Also B - A = {s | s belongs to B and s does not belong to A} = {4}. Additionally, A intersect B = {s | s belongs to A and s also belongs to B} = {1, 2}. Can we then conjecture that A = (A Ո B) U (A - B) and B = (B Ո A) U (B - A)?
Thank u this was very helpful 👍
Glad to hear it! Thanks for watching!
Thank you so much ❤
My pleasure!
I have a fun problem for you : Let G be a simple graph with 3n² vertices (n≥2). It is known that the degree of each vertex of G is not greater than 4n, there exists at least a vertex of degree one, and between any two vertices, there is a path of length ≤3. Prove that the minimum number of edges that G might have is equal to( 7n² - 3n)/2.
Hope you find it interesting.
How do you solve A U (G\M)
Thanks for watching and I am not sure what you mean. Assuming that G and M are sets, A U (G\M) is just a set, so what is the problem? What is being asked? Do we know what A, G, and M are, and want to find A U (G\M)?
God bless you!
If A is subset of B then what is B'-A'?
Thanks for watching! That would be the empty set - try to prove it! The intuition is that since A is a subset of B, it contains less than B, which means A' contains MORE than B', so taking A' away from B' would get rid of everything, leaving the empty set.
Given sets A, B, C, and U, find the elements in AB'.
A{0,1,3 }
B{1, 3,6}
C{ 5,7}
U{0, 1, 2, 3, 4, 5, 6, 7}
Let us use inductive reasoning and start backward from the end result. Let us assume that A = {1, 2, 3} and B = {1, 2, 4}. Therefore A - B = {s | s belongs to A and s does not belong to B} = {3}. Also B - A = {s | s belongs to B and s does not belong to A} = {4}. Additionally, A intersect B = {s | s belongs to A and s also belongs to B} = {1, 2}.
Can we then conjecture that A = (A Ո B) U (A - B) and B = (B Ո A) U (B - A)?